How can I solve the following inequality algebraicly for $x$?
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot (\frac{x}{50})^\frac{5}{2}\cdot e^\sqrt{\frac{x}{50t}}<x\ \ \ \ \ \ \ (t>0)$$
I suppose it can be solved with using Lambert W function but I don't know how I can introduce Lambert W here.
$\endgroup$ 12 Answers
$\begingroup$$$\frac{\sqrt{5}}{\sqrt{t}}\cdot (\frac{x}{50})^\frac{5}{2}\cdot e^\sqrt{\frac{x}{50t}}<x\ \ \ \left(t,x\in\mathbb{R}\right)$$
Reshape your inequality into an inequality of only one function. Bring for that all $x$ to the left-hand side of your inequality:
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{5}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}-x<0$$
or
$$x<0:\ \frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{3}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}>1$$
$$x>0:\ \frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{3}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}<1$$
Consider the inequalities later and solve now one of the equations that are related to this inequalities:
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{5}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}-x=0$$
or
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{3}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}=1$$
On the left-hand side of this equation, you have a function in dependence of $x$. This function is an algebraic function in dependence of $x$ and $e^{\sqrt{x}}$. $x$ and $e^{\sqrt{x}}$ are algebraically independent for all $x\neq 0$. But $x=0$ is not a solution of your inequality. Therefore this function cannot be brought into a form of an algebraic function in dependence of only one transcendental argument. Therefore you cannot solve the equation by only applying elementary operations/functions.
Lambert W function is the inverse of the function $f\colon x\mapsto f(x)=xe^x.$ For applying only Lambert W and elementary functions, your equation should be in the form
$$f_1(f_2(x)e^{f_2(x)})=c,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
where $c$ constant and $f_1$ and $f_2$ are elementary functions with a suitable elementary partial inverse.
Collect all powers of $x$ and all powers of $e^{x}$:
$$\frac{\sqrt{5}}{\sqrt{t}\cdot 50^{\frac{5}{2}}}\cdot x^\frac{5}{2}\cdot e^{\frac{x^{\frac{1}{2}}}{\sqrt{50t}}}-x=0\ \ \ \ \ \ \ \ \ (2)$$
or
$$\frac{\sqrt{5}}{\sqrt{t}\cdot 50^{\frac{5}{2}}}\cdot x^\frac{3}{2}\cdot e^{\frac{x^{\frac{1}{2}}}{\sqrt{50t}}}=1.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$
Solve one of this equations for $x$ by applying Lambert W. Both are algebraically solvable by Lambert W.
Take e.g. equation (3):
$$a=\frac{\sqrt{5}}{\sqrt{t}\cdot 50^{\frac{5}{2}}}\in\mathbb{R},\ b=\sqrt{50t}\in\mathbb{R},\ x\in\mathbb{R}$$
$$ax^\frac{3}{2}e^{\frac{x^{\frac{1}{2}}}{b}}=1$$
$$z=x^{\frac{1}{2}}:$$
$$az^{3}e^{\frac{z}{b}}=1\ \ |\ ()^{\frac{1}{3}}$$
$$a^\frac{1}{3}ze^{\frac{z}{3b}}=1\ \ |\ :3a^{\frac{1}{3}}b$$
$$\frac{z}{3b}e^{\frac{z}{3b}}=\frac{1}{3a^{\frac{1}{3}}b}$$
$$\frac{z}{3b}=W\left(\frac{1}{3a^{\frac{1}{3}}b}\right)$$
$$z=3b\cdot W\left(\frac{1}{3a^{\frac{1}{3}}b}\right)$$
$$x=z^{2}:$$
$$x=9b^{2}\cdot W\left(\frac{1}{3a^{\frac{1}{3}}b}\right)^{2}$$
Solve the above inequalities then.
$\endgroup$ $\begingroup$The solution is of the form $x\in(0, x^*(t))$ for some $x^*(t)>0$.
To see this, first notice that the solution is restricted to $\mathbb R_+$. Let $$f(x;t)\equiv\sqrt{\frac{5}{t}}\left(\frac{x}{50}\right)^{\frac 52}e^{\sqrt{\frac{x}{50t}}},\,\,\,\,g(x)\equiv x$$ Then note that $f(0;t)=g(0)=0$, while $f(x;t)$ is strictly convex, as $$f_x(x;t)=e^{\sqrt{\frac{x}{50t}}}\left(\frac 52x^{\frac 32}+\frac{x^2}{2}\sqrt{\frac{1}{50t}}\right)\text{ is strictly increasing}$$ and $\lim_{x\to+\infty}f_x(x;t)=+\infty$. So except for $0$, there exists some $x^*(t)>0$ at which $f(x^*(t);t)=x^*(t)$, and $f(x;t)<x$ on $(0,x^*(t))$ while $f(x;t)\ge x$ on $[x^*(t),\infty)$.
However, I do not think there is a closed-form characterization for $x^*(t)$.
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